The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
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The force constant of the spring is determined as 14,222.2 N/m.
<h3>Force constant of the spring</h3>
Apply the principle of conservation of energy,
K.E = U
where;
- K.E kinetic energy of the elevator
- U is elastic potential energy of the spring
¹/₂mv² = ¹/₂kx²
mv² = kx²
k = mv²/x²
Where;
- m is mass of the elevator
- v is speed
- x is compression of the spring
k = (2000 x 8²)/(3²)
k = 14,222.2 N/m
Thus, the force constant of the spring is determined as 14,222.2 N/m.
Learn more about force constant here: brainly.com/question/1968517
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Answer:
Hindi ko po alam
Explanation:
Kaya sa iba ka nalang mag tanong kac Bobo ako jaan
The average density of the boat, including the steel and air, is less dense than 1.00 g/cm³.